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a, \(5\sqrt{2x^2+3x+9}=2x^2+3x+3\) (*)
Đặt \(2x^2+3x=a\left(a\ge-9\right)\)
=> \(5\sqrt{a+9}=a+3\)
<=> \(25\left(a+9\right)=a^2+6a+9\)
<=> \(25a+225=a^2+6a+9\)
<=> \(0=a^2+6a+9-25a-225=a^2-19a-216\)
<=> 0= \(a^2-27a+8a-216\)
<=> \(\left(a-27\right)\left(a+8\right)=0\)
=> \(\left[{}\begin{matrix}a=27\\a=-8\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}2x^2+3x=27\\2x^2+3x=-8\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}2x^2+3x-27=0\\2x^2+3x+8=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\left(x-3\right)\left(2x+9\right)=0\\2\left(x^2+2.\frac{3}{4}+\frac{9}{16}\right)+\frac{55}{8}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\frac{9}{2}\left(tm\right)\\2\left(x+\frac{3}{4}\right)^2=-\frac{55}{8}\left(ktm\right)\end{matrix}\right.\)
Vậy pt (*) có tập nghiệm \(S=\left\{3,-\frac{9}{2}\right\}\)
b, \(9-\sqrt{81-7x^3}=\frac{x^3}{2}\left(đk:x\le\sqrt[3]{\frac{81}{7}}\right)\)(*)
<=> \(\sqrt{81-7x^3}=9-\frac{x^3}{2}\)
<=>\(81-7x^3=\left(9-\frac{x^3}{2}\right)^2=81-9x^3+\frac{x^6}{4}\)
<=> \(-7x^3+9x^3-\frac{x^6}{4}=0\) <=> \(2x^3-\frac{x^6}{4}=0\)<=> \(8x^3-x^6=0\)
<=> \(x^3\left(8-x^2\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\8=x^2\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=\pm2\sqrt{2}\left(ktm\right)\end{matrix}\right.\)
Vậy pt (*) có nghiệm x=0
d,\(\sqrt{9x-2x^2}-9x+2x^2+6=0\) (*) (đk: \(0\le x\le\frac{1}{2}\))
<=> \(\sqrt{9x-2x^2}-\left(9x-2x^2\right)+6=0\)
Đặt \(\sqrt{9x-2x^2}=a\left(a\ge0\right)\)
Có \(a-a^2+6=0\)
<=> \(a^2-a-6=0\) <=> \(a^2-3x+2x-6=0\)
<=> \(\left(a-3\right)\left(a+2\right)=0\)
=> \(a-3=0\) (vì a+2>0 vs mọi \(a\ge0\))
<=> a=3 <=>\(\sqrt{9x-2x^2}=3\) <=> \(9x-2x^2=9\)
<=> 0=\(2x^2-9x+9\) <=> \(2x^2-6x-3x+9=0\) <=>\(\left(2x-3\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}2x=3\\x=3\end{matrix}\right.< =>\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)(t/m)
Vậy pt (*) có tập nghiệm \(S=\left\{\frac{3}{2},3\right\}\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
a, ĐKXĐ : Tự tìm hộ hen :)
Ta có : \(\sqrt{2x^2-9x+4}+3\sqrt{2x-1}=\sqrt{2x^2+21x-11}\)
=> \(\sqrt{2x^2-9x+4}+3\sqrt{2x-1}-\sqrt{2x^2+21x-11}=0\)
=> \(\sqrt{\left(x-4\right)\left(2x-1\right)}+3\sqrt{2x-1}-\sqrt{\left(2x-1\right)\left(x+11\right)}=0\)
=> \(\sqrt{2x-1}\left(\sqrt{x-4}+3-\sqrt{x+11}\right)=0\)
=> \(\left[{}\begin{matrix}\sqrt{2x-1}=0\\\sqrt{x-4}+3=\sqrt{x+11}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2x-1=0\\x-4+6\sqrt{x-4}+9=x+11\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2x-1=0\\6\sqrt{x-4}=6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}2x-1=0\\x-4=1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=5\end{matrix}\right.\) ( TM )
Vậy ...
b, ĐKXĐ : Tiếp tục tìm hộ nha :)
Ta có : \(\sqrt{1-x}+\sqrt{x^2-3x+2}+\left(x-2\right)\sqrt{\frac{x-1}{x-2}}=3\)
=> \(\sqrt{1-x}+\sqrt{\left(x-1\right)\left(x-2\right)}+\left(x-2\right)\sqrt{\frac{x-1}{x-2}}=3\)
=> \(\sqrt{1-x}+\sqrt{\left(1-x\right)\left(2-x\right)}+\left(x-2\right)\sqrt{\frac{1-x}{2-x}}=3\)
=> \(\sqrt{1-x}\left(1+\sqrt{2-x}+\frac{x-2}{\sqrt{2-x}}\right)=3\)
=> \(\sqrt{1-x}\left(1+\sqrt{2-x}+\frac{-\left(2-x\right)}{\sqrt{2-x}}\right)=3\)
=> \(\sqrt{1-x}\left(1+\sqrt{2-x}-\sqrt{2-x}\right)=3\)
=> \(\sqrt{1-x}=3\)
=> \(1-x=9\)
=> \(x=-8\left(TM\right)\)
Vậy ...
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
3.
ĐKXĐ: \(x\ge-1;x\ne13\)
\(\left(x+2\right)\left(\sqrt{x+1}-2\right)=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+2\right)\sqrt{x+1}-2x-4=\sqrt[3]{2x+1}-3\)
\(\Leftrightarrow\left(x+1\right)\sqrt{x+1}+x+1-\left(2x+1\right)-\sqrt[3]{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt[3]{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-b^3-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{x+1}=\sqrt[3]{2x+1}\) (\(x\ge-\frac{1}{2}\))
\(\Leftrightarrow\left(x+1\right)^3=\left(2x+1\right)^2\)
\(\Leftrightarrow x=?\)
2.
ĐKXĐ: \(x\ge-\frac{1}{2}\)
\(\Leftrightarrow8x^3+2x-\left(2x+2\right)\sqrt{2x+1}=0\)
Đặt \(\left\{{}\begin{matrix}2x=a\\\sqrt{2x+1}=b\end{matrix}\right.\)
\(\Rightarrow a^3+a-\left(b^2+1\right)b=0\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow2x=\sqrt{2x+1}\) (\(x\ge0\))
\(\Leftrightarrow4x^2=2x+1\)
\(\Leftrightarrow x=?\)
a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)
\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)
\(\Leftrightarrow25x-4x=-8-75\)
\(\Leftrightarrow21x=-83\)
hay \(x=-\dfrac{83}{21}\)
b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)
\(\Leftrightarrow\left|2x-1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)
\(\Leftrightarrow\left|2x+1\right|=3x-5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)
d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)
\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)
\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)
\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)
\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)
\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)
\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)
\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)
\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)
vậy: Phương trình vô nghiệm
\(\Leftrightarrow x^2-1+1-\sqrt{2x^2-3x+2}-\frac{3}{2}\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)+\frac{\left(2x-1\right)\left(x-1\right)}{1+\sqrt{2x^2-3x+2}}-\frac{3}{2}\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\frac{1}{2}+\frac{2\left(x-\frac{1}{2}\right)}{1+\sqrt{2x^2-3x+2}}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\frac{1}{2}\right)\left(1+\frac{2}{1+\sqrt{2x^2-3x+2}}\right)=0\)
Do \(\left(1+\frac{2}{1+\sqrt{2x^2-3x+2}}\right)>0\left(\forall x\right)\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(\Leftrightarrow2x^2-9x+9-3+\sqrt{9x-2x^2}=0\)
\(\Leftrightarrow2x\left(x-3\right)-3\left(x-3\right)+\frac{\left(x-3\right)\left(-2x+3\right)}{\sqrt{9x-2x^2}+3}=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x-3-\frac{2x-3}{\sqrt{9x-2x^2}+3}\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x-3\right)\left(1-\frac{1}{\sqrt{9x-2x^2}+3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{3}{2}\end{cases}}\)
TH còn lại loại bạn tự giải nha
a) đK:\(2x^2-3x+2\ge0\)
\(x^2+3-\sqrt{2x^2-3x+2}=\frac{3}{2}\left(x+1\right)\)
<=> \(2x^2+6-2\sqrt{2x^2-3x+2}=3\left(x+1\right)\)
<=> \(2x^2-3x+3-2\sqrt{2x^2-3x+2}=0\)
Đặt: \(t=\sqrt{2x^2-3x+2}\left(t\ge0\right)\)
Ta có phương trình:
\(t^2-2+3-2t=0\Leftrightarrow t^2-2t+1=0\Leftrightarrow t=1\)
Với t=1 ta có phương trình:
\(\sqrt{2x^2-3x+2}=1\Leftrightarrow2x^2-3x+1=0\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\x=\frac{1}{2}\left(tm\right)\end{cases}}\)
Vậy:...
Câu b tương tự.